Taylor-like bound for the matrix exponential tr(exp(X))

Question

I am trying to prove the following Taylor-like bound for the the matrix function $F(X) := \operatorname{tr}(\exp(X))$, and symmetric matrices $X, V$ with $\lVert V \rVert \leq 1/2$: $$ F(X+V) \leq F(X) + \nabla F(X)[V] + 2\nabla^2F(X)[V, V]. $$ Just to be clear, here are the expressions for the directional derivatives from above (I hope that they are correct): \begin{align*} \nabla F(X) &= \exp(X) \\ \nabla F(X)[V] &= \operatorname{tr}(\exp(X) V) \\ \nabla^2 F(X)[V, V] &= \sum_{n=0}^\infty \frac{1}{(n+1)!} \sum_{i=0}^{n} \operatorname{tr}(X^i V X^{n-i} V) \end{align*} My proof attempt is as follows: The multivariate Taylor theorem states that there exists $t \in [0, 1]$ such that $$ F(X+V) = F(X) + \nabla F(X)[V] + \frac12 \nabla^2F(X+tV)[V, V]. $$ Therefore, I have to prove that $$ \nabla^2F(X+tV)[V, V] \leq 4 \nabla^2F(X)[V, V]. $$ (numerical experiments suggest that the constant 4 can be replaced with $\sqrt{e}$). It seems to me that it is easy to bound the LHS from above, but I don't see any way of bounding the RHS from below. Does anyone have a pointer or a reference for proving this?

Answer 1

I realized that I am actually better off proving a related statement that is easier to use later on $$ \operatorname{tr}(e^{X+V}) \leq \operatorname{tr}(e^X) + \operatorname{tr}(e^X V) + 2 \operatorname{tr}(e^X V^2). $$ If $X$ and $V$ commute, this is equivalent to the original statement.

We start by recalling the Golden-Thompson inequality: $\operatorname{tr}(e^{X+V}) \leq \operatorname{tr}(e^X e^V)$ for Hermitian matrices $X$ and $V$. Then, we develop $e^V$ into series and obtain \begin{align*} \operatorname{tr}(e^{X+V}) &\leq \operatorname{tr}(e^X e^V) = \sum_{n=0}^\infty \frac{1}{n!} \operatorname{tr}(e^X V^n) \\ &= \operatorname{tr}(e^X) + \operatorname{tr}(e^X V) + \sum_{n=2}^{\infty} \frac{1}{n!}\operatorname{tr}(e^X V^n) \\ &\leq \operatorname{tr}(e^X) + \operatorname{tr}(e^X V) + \sum_{n=0}^{\infty} \frac{1}{n!}\operatorname{tr}(e^X V^2 |V|^n), \end{align*} where $|V|$ is obtained from $V$ by replacing its eigenvalues with their absolute values (i.e. singular values of $V$). We handle the last term by using von Neumann's trace inequality for PSD matrices: \begin{align*} \sum_{n=0}^{\infty} \frac{1}{n!}\operatorname{tr}(e^X V^2 |V|^n) &\leq \sum_{n=0}^{\infty} \frac{1}{n!} \sum_i \lambda_i(e^X V^2) \lambda_i(|V|^n) \\ &\leq \sum_{n=0}^{\infty} \frac{1}{n!} \sum_i \lambda_i(e^X V^2) \lVert V \rVert^n = e^{\lVert V \rVert} \operatorname{tr}(e^X V^2). \end{align*} We conclude by noting that $e^{\lVert V \rVert} \leq \sqrt{e} \leq 2$.